Information about the test:
| question | description |
|---|---|
| A1 | linear function |
| A2 | 3D |
| A3 | arithmetic rules |
| A4 | Easy ineq. |
| A5 | logs |
| A6 | logs |
| A7 | graph translation |
| A8 | sine pi/3 |
| A9 | trig.ineq. |
| A10 | trig.identity |
| A11 | period |
| A12 | rational exponents |
| A13 | ellipsoid |
| A14 | limit |
| A15 | series |
| A16 | diff.quotient |
| A17 | graph f' |
| A18 | product rule |
| A19 | quotient rule |
| A20 | (exp)' |
| A21 | (ln(sin))' |
| A22 | hyp.functions |
| A23 | slope tangent |
| A24 | IVT |
| A25 | velocity |
| A26 | int(poly) |
| A27 | int(exp) |
| A28 | Int = 0 |
| A29 | int even funct |
| A31 | int(abs) |
| A32 | FtoC algebra |
| A33 | difference vectors |
| A35 | intersect planes |
| A36 | parallel planes |
| A30 | FtoC graph |
| A34 | normal vector |
Load the student scores for the test - here we load the 2017 and 2018 ETH Zurich test data:
test_scores
## # A tibble: 3,433 x 38
## year class A1 A2 A3 A4 A5 A6 A7 A8 A9 A10 A11
## <chr> <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 2017 s21t~ 1 0 1 1 1 0 1 0 1 0 1
## 2 2017 s21t~ 1 0 1 1 1 1 0 1 1 2 1
## 3 2017 s21t~ 1 0 0 0 1 1 1 0 1 1 1
## 4 2017 s21t~ 1 0 1 1 1 1 1 1 1 1 1
## 5 2017 s21t~ 1 0 1 0 2 0 1 0 1 0 2
## 6 2017 s21t~ 0 1 0 0 1 2 0 2 2 2 1
## 7 2017 s21t~ 1 0 1 0 2 1 0 1 2 1 0
## 8 2017 s21t~ 1 1 1 1 2 1 1 2 2 2 2
## 9 2017 s21t~ 1 1 0 1 1 1 1 1 1 1 1
## 10 2017 s21t~ 1 0 1 0 0 1 0 0 1 1 1
## # ... with 3,423 more rows, and 25 more variables: A12 <dbl>, A13 <dbl>,
## # A14 <dbl>, A15 <dbl>, A16 <dbl>, A17 <dbl>, A18 <dbl>, A19 <dbl>,
## # A20 <dbl>, A21 <dbl>, A22 <dbl>, A23 <dbl>, A24 <dbl>, A25 <dbl>,
## # A26 <dbl>, A27 <dbl>, A28 <dbl>, A29 <dbl>, A30 <dbl>, A31 <dbl>,
## # A32 <dbl>, A33 <dbl>, A34 <dbl>, A35 <dbl>, A36 <dbl>
The data includes scores of 2 for the “I don’t know” answer option. We replace these with NA:
test_scores <- test_scores %>%
mutate(across(starts_with("A"), ~ na_if(., 2))) %>%
# A few students gave "I don't know" as their answer to every question
# Here we remove them, since rows with all NAs cause problems for mirt
filter(!if_all(starts_with("A"), is.na))
The number of responses from each class:
test_scores %>%
group_by(year, class) %>%
tally() %>%
gt() %>%
data_color(
columns = c("n"),
colors = scales::col_numeric(palette = c("Blues"), domain = NULL)
)
## Warning: The `.dots` argument of `group_by()` is deprecated as of dplyr 1.0.0.
| class | n |
|---|---|
| 2017 | |
| s21t-000-2017 | 1678 |
| 2018 | |
| s21t-000-2018 | 1746 |
Mean and standard deviation for each item:
test_scores %>%
select(-class) %>%
group_by(year) %>%
skim_without_charts() %>%
select(-contains("character."), -contains("numeric.p"), -skim_type) %>%
group_by(year) %>%
gt() %>%
fmt_number(columns = contains("numeric"), decimals = 3) %>%
data_color(
columns = c("numeric.mean"),
colors = scales::col_numeric(palette = c("Greens"), domain = NULL)
) %>%
cols_label(
numeric.mean = "Mean",
numeric.sd = "SD"
)
| skim_variable | n_missing | complete_rate | Mean | SD |
|---|---|---|---|---|
| 2017 | ||||
| A1 | 15 | 0.9910608 | 0.960 | 0.195 |
| A2 | 131 | 0.9219309 | 0.333 | 0.471 |
| A3 | 77 | 0.9541120 | 0.655 | 0.476 |
| A4 | 27 | 0.9839094 | 0.655 | 0.476 |
| A5 | 544 | 0.6758045 | 0.692 | 0.462 |
| A6 | 362 | 0.7842670 | 0.901 | 0.298 |
| A7 | 19 | 0.9886770 | 0.716 | 0.451 |
| A8 | 311 | 0.8146603 | 0.583 | 0.493 |
| A9 | 613 | 0.6346841 | 0.731 | 0.444 |
| A10 | 452 | 0.7306317 | 0.745 | 0.436 |
| A11 | 352 | 0.7902265 | 0.729 | 0.445 |
| A12 | 216 | 0.8712753 | 0.784 | 0.412 |
| A13 | 211 | 0.8742551 | 0.429 | 0.495 |
| A14 | 237 | 0.8587604 | 0.653 | 0.476 |
| A15 | 269 | 0.8396901 | 0.551 | 0.498 |
| A16 | 427 | 0.7455304 | 0.249 | 0.433 |
| A17 | 154 | 0.9082241 | 0.649 | 0.477 |
| A18 | 468 | 0.7210965 | 0.529 | 0.499 |
| A19 | 566 | 0.6626937 | 0.533 | 0.499 |
| A20 | 108 | 0.9356377 | 0.776 | 0.417 |
| A21 | 408 | 0.7568534 | 0.674 | 0.469 |
| A22 | 477 | 0.7157330 | 0.723 | 0.448 |
| A23 | 492 | 0.7067938 | 0.584 | 0.493 |
| A24 | 326 | 0.8057211 | 0.730 | 0.444 |
| A25 | 78 | 0.9535161 | 0.568 | 0.495 |
| A26 | 250 | 0.8510131 | 0.938 | 0.241 |
| A27 | 590 | 0.6483909 | 0.603 | 0.490 |
| A28 | 647 | 0.6144219 | 0.686 | 0.464 |
| A29 | 373 | 0.7777116 | 0.624 | 0.485 |
| A30 | 145 | 0.9135876 | 0.830 | 0.376 |
| A31 | 214 | 0.8724672 | 0.432 | 0.495 |
| A32 | 466 | 0.7222884 | 0.315 | 0.465 |
| A33 | 57 | 0.9660310 | 0.808 | 0.394 |
| A34 | 497 | 0.7038141 | 0.826 | 0.379 |
| A35 | 886 | 0.4719905 | 0.688 | 0.464 |
| A36 | 830 | 0.5053635 | 0.436 | 0.496 |
| 2018 | ||||
| A1 | 16 | 0.9908362 | 0.964 | 0.186 |
| A2 | 97 | 0.9444444 | 0.300 | 0.458 |
| A3 | 37 | 0.9788087 | 0.658 | 0.475 |
| A4 | 23 | 0.9868270 | 0.634 | 0.482 |
| A5 | 441 | 0.7474227 | 0.659 | 0.474 |
| A6 | 269 | 0.8459336 | 0.870 | 0.336 |
| A7 | 13 | 0.9925544 | 0.711 | 0.453 |
| A8 | 199 | 0.8860252 | 0.561 | 0.496 |
| A9 | 549 | 0.6855670 | 0.673 | 0.469 |
| A10 | 367 | 0.7898053 | 0.687 | 0.464 |
| A11 | 293 | 0.8321879 | 0.700 | 0.458 |
| A12 | 127 | 0.9272623 | 0.770 | 0.421 |
| A13 | 188 | 0.8923253 | 0.401 | 0.490 |
| A14 | 192 | 0.8900344 | 0.645 | 0.479 |
| A15 | 220 | 0.8739977 | 0.530 | 0.499 |
| A16 | 341 | 0.8046964 | 0.250 | 0.433 |
| A17 | 120 | 0.9312715 | 0.648 | 0.478 |
| A18 | 372 | 0.7869416 | 0.463 | 0.499 |
| A19 | 468 | 0.7319588 | 0.468 | 0.499 |
| A20 | 85 | 0.9513173 | 0.762 | 0.426 |
| A21 | 336 | 0.8075601 | 0.646 | 0.478 |
| A22 | 420 | 0.7594502 | 0.682 | 0.466 |
| A23 | 397 | 0.7726231 | 0.558 | 0.497 |
| A24 | 260 | 0.8510882 | 0.678 | 0.467 |
| A25 | 46 | 0.9736541 | 0.753 | 0.431 |
| A26 | 230 | 0.8682703 | 0.929 | 0.256 |
| A27 | 531 | 0.6958763 | 0.576 | 0.494 |
| A28 | 529 | 0.6970218 | 0.655 | 0.476 |
| A29 | 319 | 0.8172967 | 0.633 | 0.482 |
| A30 | 121 | 0.9306987 | 0.786 | 0.410 |
| A31 | 166 | 0.9049255 | 0.411 | 0.492 |
| A32 | 436 | 0.7502864 | 0.272 | 0.445 |
| A33 | 63 | 0.9639175 | 0.791 | 0.407 |
| A34 | 425 | 0.7565865 | 0.815 | 0.389 |
| A35 | 797 | 0.5435281 | 0.612 | 0.488 |
| A36 | 741 | 0.5756014 | 0.397 | 0.490 |
Before applying IRT, we should check that the data satisfies the assumptions needed by the model. In particular, to use a 1-dimensional IRT model, we should have some evidence of unidimensionality in the test scores.
TODO Perhaps just remove this section? See the later section on local independence which is consistent with DeMars p.48
This plot shows the correlations between scores on each pair of items:
item_scores <- test_scores %>%
select(-class, -year)
cor_ci <- psych::corCi(item_scores, plot = FALSE)
psych::cor.plot.upperLowerCi(cor_ci)
There are a few correlations that are not significantly different from 0:
cor_ci$ci %>%
as_tibble(rownames = "corr") %>%
filter(p > 0.05) %>%
arrange(-p) %>%
select(-contains(".e")) %>%
gt() %>%
fmt_number(columns = 2:4, decimals = 3)
| corr | lower | upper | p |
|---|---|---|---|
| A1-A29 | −0.024 | 0.047 | 0.519 |
| A24-A29 | −0.023 | 0.055 | 0.417 |
| A2-A34 | −0.021 | 0.062 | 0.339 |
| A29-A31 | −0.062 | 0.021 | 0.331 |
| A1-A2 | −0.014 | 0.051 | 0.260 |
| A1-A5 | −0.014 | 0.069 | 0.194 |
| A17-A29 | −0.010 | 0.071 | 0.146 |
| A2-A25 | −0.009 | 0.067 | 0.139 |
| A13-A29 | −0.007 | 0.078 | 0.099 |
| A2-A29 | −0.007 | 0.079 | 0.098 |
| A1-A28 | −0.007 | 0.086 | 0.097 |
| A1-A12 | −0.003 | 0.069 | 0.073 |
| A16-A34 | −0.003 | 0.084 | 0.070 |
| A29-A30 | −0.001 | 0.079 | 0.054 |
The overall picture is that the item scores are well correlated with each other.
structure <- check_factorstructure(item_scores)
n <- n_factors(item_scores)
The choice of 1 dimensions is supported by 4 (17.39%) methods out of 23 (Acceleration factor, VSS complexity 1, Velicer’s MAP, RMSEA).
plot(n)
summary(n) %>% gt()
| n_Factors | n_Methods |
|---|---|
| 1 | 4 |
| 2 | 1 |
| 3 | 1 |
| 4 | 4 |
| 5 | 1 |
| 7 | 2 |
| 12 | 2 |
| 18 | 1 |
| 26 | 1 |
| 28 | 1 |
| 33 | 1 |
| 34 | 1 |
| 35 | 3 |
#n %>% tibble() %>% gt()
fa.parallel(item_scores, fa = "fa")
## Parallel analysis suggests that the number of factors = 11 and the number of components = NA
We use the factanal function to fit a 1-factor model. Note that this function cannot handle missing data, so we set the NA scores to 0 for this analysis.
fitfact <- factanal(item_scores %>%
mutate(across(everything(), ~ replace_na(.x, 0))),
factors = 1,
rotation = "varimax")
print(fitfact, digits = 2, cutoff = 0.3, sort = TRUE)
##
## Call:
## factanal(x = item_scores %>% mutate(across(everything(), ~replace_na(.x, 0))), factors = 1, rotation = "varimax")
##
## Uniquenesses:
## A1 A2 A3 A4 A5 A6 A7 A8 A9 A10 A11 A12 A13 A14 A15 A16
## 0.96 0.95 0.76 0.80 0.72 0.75 0.79 0.79 0.61 0.66 0.68 0.88 0.91 0.73 0.87 0.81
## A17 A18 A19 A20 A21 A22 A23 A24 A25 A26 A27 A28 A29 A30 A31 A32
## 0.78 0.78 0.78 0.64 0.59 0.61 0.64 0.73 0.95 0.69 0.63 0.65 0.85 0.81 0.79 0.78
## A33 A34 A35 A36
## 0.89 0.86 0.79 0.84
##
## Loadings:
## [1] 0.53 0.50 0.62 0.59 0.57 0.52 0.60 0.64 0.62 0.60 0.52 0.55 0.61 0.60
## [16] 0.49 0.45 0.45 0.46 0.35 0.31 0.36 0.44 0.47 0.46 0.47 0.38 0.43
## [31] 0.46 0.47 0.33 0.37 0.46 0.40
##
## Factor1
## SS loadings 8.27
## Proportion Var 0.23
##
## Test of the hypothesis that 1 factor is sufficient.
## The chi square statistic is 5393.45 on 594 degrees of freedom.
## The p-value is 0
load <- tidy(fitfact)
ggplot(load, aes(x = fl1, y = 0)) +
geom_point() +
geom_label_repel(aes(label = paste0("A", rownames(load))), show.legend = FALSE) +
labs(x = "Factor 1", y = NULL,
title = "Standardised Loadings",
subtitle = "Based upon correlation matrix") +
theme_minimal()
## Warning: ggrepel: 15 unlabeled data points (too many overlaps). Consider
## increasing max.overlaps
load %>%
select(question = variable, factor_loading = fl1) %>%
left_join(item_info, by = "question") %>%
gt() %>%
data_color(
columns = contains("factor"),
colors = scales::col_numeric(palette = c("Greens"), domain = NULL)
)
| question | factor_loading | description |
|---|---|---|
| A1 | 0.2094343 | linear function |
| A2 | 0.2320310 | 3D |
| A3 | 0.4938949 | arithmetic rules |
| A4 | 0.4507755 | Easy ineq. |
| A5 | 0.5309866 | logs |
| A6 | 0.5041101 | logs |
| A7 | 0.4543472 | graph translation |
| A8 | 0.4582495 | sine pi/3 |
| A9 | 0.6222243 | trig.ineq. |
| A10 | 0.5862856 | trig.identity |
| A11 | 0.5687470 | period |
| A12 | 0.3457532 | rational exponents |
| A13 | 0.3051984 | ellipsoid |
| A14 | 0.5208039 | limit |
| A15 | 0.3575489 | series |
| A16 | 0.4372204 | diff.quotient |
| A17 | 0.4670996 | graph f' |
| A18 | 0.4636871 | product rule |
| A19 | 0.4712863 | quotient rule |
| A20 | 0.6026572 | (exp)' |
| A21 | 0.6398888 | (ln(sin))' |
| A22 | 0.6213857 | hyp.functions |
| A23 | 0.5995268 | slope tangent |
| A24 | 0.5187729 | IVT |
| A25 | 0.2305446 | velocity |
| A26 | 0.5537627 | int(poly) |
| A27 | 0.6075910 | int(exp) |
| A28 | 0.5956103 | Int = 0 |
| A29 | 0.3846979 | int even funct |
| A30 | 0.4319645 | FtoC graph |
| A31 | 0.4631794 | int(abs) |
| A32 | 0.4687184 | FtoC algebra |
| A33 | 0.3343483 | difference vectors |
| A34 | 0.3725708 | normal vector |
| A35 | 0.4572324 | intersect planes |
| A36 | 0.4011620 | parallel planes |
# To proceed with the IRT analysis, comment out the following line before knitting
#knitr::knit_exit()
We can fit a Multidimensional Item Response Theory (mirt) model. From the function definition:
mirt fits a maximum likelihood (or maximum a posteriori) factor analysis model to any mixture of dichotomous and polytomous data under the item response theory paradigm using either Cai's (2010) Metropolis-Hastings Robbins-Monro (MHRM) algorithm.
The process is to first fit the model, and save the result as a model object that we can then parse to get tabular or visual displays of the model that we might want. When fitting the model, we have the option to specify a few arguments, which then get interpreted as parameters to be passed to the model.
fit_2pl <- mirt(
data = item_scores, # just the columns with question scores
model = 1, # number of factors to extract
itemtype = "2PL", # 2 parameter logistic model
SE = TRUE # estimate standard errors
)
##
Iteration: 1, Log-Lik: -55174.300, Max-Change: 1.11769
Iteration: 2, Log-Lik: -53440.531, Max-Change: 0.24450
Iteration: 3, Log-Lik: -53280.158, Max-Change: 0.12660
Iteration: 4, Log-Lik: -53207.382, Max-Change: 0.08141
Iteration: 5, Log-Lik: -53164.273, Max-Change: 0.05506
Iteration: 6, Log-Lik: -53136.182, Max-Change: 0.04470
Iteration: 7, Log-Lik: -53116.411, Max-Change: 0.03939
Iteration: 8, Log-Lik: -53102.191, Max-Change: 0.03652
Iteration: 9, Log-Lik: -53091.778, Max-Change: 0.03240
Iteration: 10, Log-Lik: -53084.093, Max-Change: 0.02883
Iteration: 11, Log-Lik: -53078.411, Max-Change: 0.02504
Iteration: 12, Log-Lik: -53074.180, Max-Change: 0.02185
Iteration: 13, Log-Lik: -53061.970, Max-Change: 0.00565
Iteration: 14, Log-Lik: -53061.822, Max-Change: 0.00437
Iteration: 15, Log-Lik: -53061.712, Max-Change: 0.00387
Iteration: 16, Log-Lik: -53061.388, Max-Change: 0.00164
Iteration: 17, Log-Lik: -53061.378, Max-Change: 0.00124
Iteration: 18, Log-Lik: -53061.374, Max-Change: 0.00105
Iteration: 19, Log-Lik: -53061.364, Max-Change: 0.00029
Iteration: 20, Log-Lik: -53061.364, Max-Change: 0.00018
Iteration: 21, Log-Lik: -53061.364, Max-Change: 0.00018
Iteration: 22, Log-Lik: -53061.362, Max-Change: 0.00016
Iteration: 23, Log-Lik: -53061.362, Max-Change: 0.00014
Iteration: 24, Log-Lik: -53061.362, Max-Change: 0.00012
Iteration: 25, Log-Lik: -53061.362, Max-Change: 0.00007
##
## Calculating information matrix...
We compute Yen’s \(Q_3\) (1984) to check for any dependence between items after controlling for \(\theta\). This gives a score for each pair of items, with scores above 0.2 regarded as problematic (see DeMars, p. 48).
residuals_2pl %>% as.matrix() %>%
corrplot::corrplot(type = "upper")
This shows that most item pairs are independent, with only one pair showing cause for concern:
residuals_2pl %>%
rownames_to_column(var = "q1") %>%
as_tibble() %>%
pivot_longer(cols = starts_with("A"), names_to = "q2", values_to = "Q3_score") %>%
filter(abs(Q3_score) > 0.2) %>%
filter(parse_number(q1) < parse_number(q2)) %>%
gt()
| q1 | q2 | Q3_score |
|---|---|---|
| A18 | A19 | 0.6691425 |
Items A18 and A19 are on the product and quotient rules.
Given that this violation of the local independence assumption is very mild, we proceed using this model.
We then compute factor score estimates and augment the existing data frame with these estimates, to keep everything in one place. To do the estimation, we use the fscores() function from the mirt package which takes in a computed model object and computes factor score estimates according to the method specified. We will use the EAP method for factor score estimation, which is the “expected a-posteriori” method, the default. We specify it explicitly below, but the results would have been the same if we omitted specifying the method argument since it’s the default method the function uses.
test_scores <- test_scores %>%
mutate(F1 = fscores(fit_2pl, method = "EAP"))
We can also calculate the model coefficient estimates using a generic function coef() which is used to extract model coefficients from objects returned by modeling functions. We will set the IRTpars argument to TRUE, which means slope intercept parameters will be converted into traditional IRT parameters.
coefs_2pl <- coef(fit_2pl, IRTpars = TRUE)
The resulting object coefs is a list, with one element for each question, and an additional GroupPars element that we won’t be using. The output is a bit long, so we’re only showing a few of the elements here:
coefs_2pl[1:3]
## $A1
## a b g u
## par 0.9470642 -3.817165 0 1
## CI_2.5 0.7202175 -4.576813 NA NA
## CI_97.5 1.1739108 -3.057518 NA NA
##
## $A2
## a b g u
## par 0.6050046 1.408046 0 1
## CI_2.5 0.5136783 1.184206 NA NA
## CI_97.5 0.6963309 1.631886 NA NA
##
## $A3
## a b g u
## par 1.327727 -0.6394136 0 1
## CI_2.5 1.200583 -0.7191701 NA NA
## CI_97.5 1.454871 -0.5596570 NA NA
# coefs_2pl[35:37]
Let’s take a closer look at the first element:
coefs_2pl[1]
## $A1
## a b g u
## par 0.9470642 -3.817165 0 1
## CI_2.5 0.7202175 -4.576813 NA NA
## CI_97.5 1.1739108 -3.057518 NA NA
In this output:
a is discriminationb is difficultyTo make this output a little more user friendly, we should tidy it such that we have a row per question. We’ll do this in two steps. First, write a function that tidies the output for one question, i.e. one list element. Then, map this function over the list of all questions, resulting in a data frame.
tidy_mirt_coefs <- function(x){
x %>%
# melt the list element
melt() %>%
# convert to a tibble
as_tibble() %>%
# convert factors to characters
mutate(across(where(is.factor), as.character)) %>%
# only focus on rows where X2 is a or b (discrimination or difficulty)
filter(X2 %in% c("a", "b")) %>%
# in X1, relabel par (parameter) as est (estimate)
mutate(X1 = if_else(X1 == "par", "est", X1)) %>%
# unite columns X2 and X1 into a new column called var separated by _
unite(X2, X1, col = "var", sep = "_") %>%
# turn into a wider data frame
pivot_wider(names_from = var, values_from = value)
}
Let’s see what this does to a single element in coefs:
tidy_mirt_coefs(coefs_2pl[1])
## # A tibble: 1 x 7
## L1 a_est a_CI_2.5 a_CI_97.5 b_est b_CI_2.5 b_CI_97.5
## <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 A1 0.947 0.720 1.17 -3.82 -4.58 -3.06
And now let’s map it over all 32 elements of coefs:
tidy_2pl <- map_dfr(coefs_2pl[1:32], tidy_mirt_coefs, .id = "Question")
A quick peek at the result:
tidy_2pl
## # A tibble: 32 x 7
## Question a_est a_CI_2.5 a_CI_97.5 b_est b_CI_2.5 b_CI_97.5
## <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 A1 0.947 0.720 1.17 -3.82 -4.58 -3.06
## 2 A2 0.605 0.514 0.696 1.41 1.18 1.63
## 3 A3 1.33 1.20 1.45 -0.639 -0.719 -0.560
## 4 A4 1.28 1.16 1.40 -0.604 -0.684 -0.525
## 5 A5 1.03 0.906 1.16 -0.634 -0.756 -0.513
## 6 A6 1.01 0.854 1.18 -2.20 -2.50 -1.90
## 7 A7 1.36 1.23 1.49 -0.904 -0.992 -0.815
## 8 A8 1.03 0.920 1.14 -0.286 -0.375 -0.197
## 9 A9 1.59 1.41 1.76 -0.489 -0.577 -0.401
## 10 A10 1.47 1.31 1.63 -0.687 -0.779 -0.596
## # ... with 22 more rows
And a nicely formatted table of the result:
gt(tidy_2pl) %>%
fmt_number(columns = contains("_"), decimals = 3) %>%
data_color(
columns = contains("a_"),
colors = scales::col_numeric(palette = c("Greens"), domain = NULL)
) %>%
data_color(
columns = contains("b_"),
colors = scales::col_numeric(palette = c("Blues"), domain = NULL)
) %>%
tab_spanner(label = "Discrimination", columns = contains("a_")) %>%
tab_spanner(label = "Difficulty", columns = contains("b_")) %>%
cols_label(
a_est = "Est.",
b_est = "Est.",
a_CI_2.5 = "2.5%",
b_CI_2.5 = "2.5%",
a_CI_97.5 = "97.5%",
b_CI_97.5 = "97.5%"
)
tidy_2pl %>%
mutate(qnum = parse_number(Question)) %>%
ggplot(aes(
x = a_est,
y = b_est
)) +
geom_errorbar(aes(ymin = b_CI_2.5, ymax = b_CI_97.5), width = 0, alpha = 0.5) +
geom_errorbar(aes(xmin = a_CI_2.5, xmax = a_CI_97.5), width = 0, alpha = 0.5) +
geom_text_repel(aes(label = Question), alpha = 0.5) +
geom_point() +
theme_minimal() +
labs(x = "Discrimination",
y = "Difficulty")
tidy_2pl %>%
write_csv("data-eth/OUT_2pl-results-pre-only.csv")
Do students from different programmes of study have different distributions of ability?
Compare the distribution of abilities in the year groups (though in this case there is only one).
ggplot(test_scores, aes(F1, fill = as.factor(year), colour = as.factor(year))) +
geom_density(alpha=0.5) +
scale_x_continuous(limits = c(-3.5,3.5)) +
labs(title = "Density plot",
subtitle = "Ability grouped by year of taking the test",
x = "Ability", y = "Density",
fill = "Year", colour = "Year") +
theme_minimal()
Compare the distribution of abilities in the various classes.
ggplot(test_scores, aes(x = F1, y = class, colour = class, fill = class)) +
geom_density_ridges(alpha = 0.5) +
scale_x_continuous(limits = c(-3.5,3.5)) +
guides(fill = FALSE, colour = FALSE) +
labs(title = "Density plot",
subtitle = "Ability grouped by class of taking the test",
x = "Ability", y = "Class") +
theme_minimal()
## Picking joint bandwidth of 0.187
plot(fit_2pl, type = "infoSE", main = "Test information")
plot(fit_2pl, type = "infotrace", main = "Item information curves")
plot(fit_2pl, type = "score", auto.key = FALSE)
We can get individual item surface and information plots using the itemplot() function from the mirt package, e.g.
mirt::itemplot(fit_2pl, item = 1,
main = "Trace lines for item 1")
We can also get the plots for all trace lines, one facet per plot.
plot(fit_2pl, type = "trace", auto.key = FALSE)
Or all of them overlaid in one plot.
plot(fit_2pl, type = "trace", facet_items=FALSE)
An alternative approach is using ggplot2 and plotly to add interactivity to make it easier to identify items.
# store the object
plt <- plot(fit_2pl, type = "trace", facet_items = FALSE)
# the data we need is in panel.args
# TODO - I had to change the [[1]] to [[2]] since the plt has two panels for some reason, with the one we want being the 2nd panel
plt_data <- tibble(
x = plt$panel.args[[2]]$x,
y = plt$panel.args[[2]]$y,
subscripts = plt$panel.args[[2]]$subscripts,
item = rep(colnames(item_scores), each = 200)
) %>%
mutate(
item_no = str_remove(item, "A") %>% as.numeric(),
item = fct_reorder(item, item_no)
)
head(plt_data)
## # A tibble: 6 x 5
## x y subscripts item item_no
## <dbl> <dbl> <int> <fct> <dbl>
## 1 -6 0.112 201 A1 1
## 2 -5.94 0.118 202 A1 1
## 3 -5.88 0.124 203 A1 1
## 4 -5.82 0.131 204 A1 1
## 5 -5.76 0.137 205 A1 1
## 6 -5.70 0.144 206 A1 1
plt_gg <- ggplot(plt_data, aes(x, y,
colour = item,
text = item)) +
geom_line() +
labs(
title = "2PL - Trace lines",
#x = expression(theta),
x = "theta",
#y = expression(P(theta)),
y = "P(theta)",
colour = "Item"
) +
theme_minimal() +
theme(legend.position = "none")
ggplotly(plt_gg, tooltip = "text")
knitr::knit_exit()